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A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. There is a unique plane in three-dimensional space which passes through the three points with these coordinates, and this plane is the set of all points whose coordinates are solutions of the equation.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.
For example, log 2 (8) = 3, because 2 3 = 8. The graph gets arbitrarily close to the y axis, but does not meet or intersect it . An exponential equation is one which has the form a x = b {\displaystyle a^{x}=b} for a > 0 {\displaystyle a>0} , [ 43 ] which has solution
The equations 3x + 2y = 6 and 3x + 2y = 12 are inconsistent. A linear system is inconsistent if it has no solution, and otherwise, it is said to be consistent. [7] When the system is inconsistent, it is possible to derive a contradiction from the equations, that may always be rewritten as the statement 0 = 1. For example, the equations
The polynomial 3x 2 − 5x + 4 is written in descending powers of x. The first term has coefficient 3, indeterminate x, and exponent 2. In the second term, the coefficient is −5. The third term is a constant. Because the degree of a non-zero polynomial is the largest degree of any one term, this polynomial has degree two. [11]
Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x (x + 1)(x + 2) – namely x = 0, x = −1, and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.